That's a nice problem! The room is 5-ft long and 4-ft wide.
If the dimensions of the rectangle are in feet, the perimeter will be in feet as well. The area will be in square feet. The area is length x width. The perimeter is 2 x (length + width) or 2 x length + 2 x width.
Area is 1.44 square feet Perimeter is 6 feet.
4 feet
2
8 by 12
If the dimensions of the rectangle are in feet, the perimeter will be in feet as well. The area will be in square feet. The area is length x width. The perimeter is 2 x (length + width) or 2 x length + 2 x width.
Area is 1.44 square feet Perimeter is 6 feet.
4 feet
4 feet
2
5
8 by 12
The width is either 2ft or 7ft.
You can't only using the perimeter. You have to know length and width.
area = width x height = 30 square feet ( not feet) 5 x height = 30 height = 6 feet perimeter = 2 width + 2 height = 10 + 12 = 22 feet
Write two simultaneous equations and solve them. One for the perimeter, one for the area.
The area of a square with a perimeter of 22.33 feet is: 31.16 square feet.